A072878 -id:A072878 - cp's OEIS frontend

A072879 a(n) = 5a(n-1)a(n-2)a(n-3)a(n-4) - a(n-5) with a(1) = a(2) = a(3) = a(4) = a(5) = 1.

1, 1, 1, 1, 1, 4, 19, 379, 144019, 20741616379, 107553662508585672001, 608831069421618273050865038881215685876, 978035016076705458999330010986670207956236476587064788804921180339451725001

Offset: 1

Benoit Cloitre, Jul 28 2002

Solutions of the Hurwitz equation in five variables.

Seiichi Manyama, Table of n, a(n) for n = 1..16
Arthur Baragar, Integral solutions of the Markoff-Hurwitz equations, J. Number Theory 49 (1994), 27-44.
Andrew N. W. Hone, Diophantine non-integrability of a third order recurrence with the Laurent property, arXiv:math/0601324 [math.NT], 2006.
Andrew N. W. Hone, Diophantine non-integrability of a third order recurrence with the Laurent property, J. Phys. A: Math. Gen. 39 (2006), L171-L177.
Matthew Christopher Russell, Using experimental mathematics to conjecture and prove theorems in the theory of partitions and commutative and non-commutative recurrences, PhD Dissertation, Mathematics Department, Rutgers University, May 2016.

Cf. A006720, A064098, A072878, A072880.

nxt[{a_,b_,c_,d_,e_}]:={b,c,d,e,(5b c d e)-a}; NestList[nxt,{1,1,1,1,1},20][[All,1]] (* Harvey P. Dale, Nov 07 2016 *)

a(1) = a(2) = a(3) = a(4) = a(5) = 1; a(n) = (a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-4)^2)/a(n-5) for n >= 6.

From the recurrence a(n) = 5*a(n-1)*a(n-2)*a(n-3)*a(n-4) - a(n-5), any five successive terms satisfy the five-variable Hurwitz equation a(n)^2+a(n-1)^2+a(n-2)^2+a(n-3)^2+a(n-4)^2 = 5*a(n)*a(n-1)*a(n-2)*a(n-3)*a(n-4). As n tends to infinity, the limit of log(log(a(n)))/n is log x = 0.6562559790..., where x=1.927561975... is the largest real root of the quartic x^4-x^3-x^2-x-1=0. - Andrew Hone, Nov 16 2005

Entry revised Nov 19 2005, based on comments from Andrew Hone

Name clarified by Petros Hadjicostas, May 11 2019

A064098 a(n+1) = (a(n)^2 + a(n-1)^2)/a(n-2), with a(1) = a(2) = a(3) = 1.

Original entry on oeis.org

1, 1, 1, 2, 5, 29, 433, 37666, 48928105, 5528778008357, 811537892743746482789, 13460438563050022083842073547074914, 32770967840592833551621556305285371426044732591005957081

Offset: 1

Santi Spadaro, Sep 16 2001

nonn

This sequence was introduced by Dana Scott but possibly studied earlier by others. - James Propp, Jan 27 2005
Sequence gives the upper-left entries of the respective matrices
  [1 1] [1 0] [2 1] [5 2] [29 12] [433 179] [37666 15571]
  [1 2] [0 1] [1 1] [2 1] [12  5] [179  74] [15571  6437], ...
satisfying the recurrence M(n) = M(n-1) M(n-3)^(-1) M(n-1) (note that the Fibonacci numbers satisfy the additive version of this recurrence). - James Propp, Jan 27 2005
Define b(1) = b(2) = b(3) = 1; b(n) = (b(n-1) + b(n-2))^2/b(n-3); then a(n) = sqrt(b(n)). - Benoit Cloitre, Jul 28 2002
Any 3 successive terms of the sequence satisfy the Markov equation x^2 + y^2 + z^2 = 3 xyz. Therefore from the 3rd term on this is a subsequence of the Markov numbers, A002559. Also, we conjecture that the limit of log(log(a(n)))/n is log((sqrt(5) + 1)/2). - Martin Giese (martin.giese(AT)oeaw.ac.at), Oct 13 2005
A subsequence of the Markoff numbers A002559. - Andrew Hone, Jan 16 2006
The recursion exhibits the Laurent phenomenon. Let F(n) = Fibonacci(n), e(n) = F(n) - 1, a(1) = a1, a(2) = a2, a(3) = a3, a(n) = (a(n-1)^2 + a(n-3)^2) / a(n-3), b(n) = a(n) * a1^e(n-1) * a2^e(n-2) * a3^e(n-3). Then b(n) for n > 1 is an irreducible polynomial in {a1^2, a2^2, a3^2}, b(n) = (b(n-1)^2 + (b(n-2) * a1^F(n-4) * a2^F(n-5) * a3^F(n-6))^2) / b(n-3), and a(n) = a(n-1) * a(n-2) * (a1^2 + a2^2 + a3^2) / (a1 * a2 * a3) - a(n-3). - Michael Somos, Jan 12 2013
Starting with n = 5, a(n) is the largest number in row n - 5 of the Markov tree, A368546. These numbers are obtained by descending the tree in alternating right and left steps; their Farey indices (see A368546 for the definition) are ratios of successive Fibonacci numbers, 1/2, 2/3, 3/5, 5/8, ... See Aigner, Proposition 10.2. - Wouter Meeussen and William P. Orrick, Feb 11 2024

			G.f. = x + x^2 + x^3 + 2*x^4 + 5*x^5 + 29*x^6 + 433*x^7 + 37666*x^8 + ...

Martin Aigner, Markov's theorem and 100 years of the uniqueness conjecture. A mathematical journey from irrational numbers to perfect matchings. Springer, 2013. x+257 pp. ISBN: 978-3-319-00887-5; 978-3-319-00888-2 MR3098784

Harry J. Smith, Table of n, a(n) for n = 1..18
Martin Aigner, Markov's theorem and 100 years of the uniqueness conjecture. A mathematical journey from irrational numbers to perfect matchings, [archive.org copy of the book].
S. Fomin and A. Zelevinsky, The Laurent Phenomenon, Advances in Applied Mathematics, 28 (2002), 119-144.
Andrew N. W. Hone, Diophantine non-integrability of a third order recurrence with the Laurent property, arXiv:math/0601324 [math.NT], 2006.
Andrew N. W. Hone, Diophantine non-integrability of a third order recurrence with the Laurent property, J. Phys. A: Math. Gen. 39 (2006), L171-L177.
Andrew N. W. Hone, Growth of Mahler measure and algebraic entropy of dynamics with the Laurent property, arXiv:2109.08217 [math.NT], 2021.
KöMaL-Mathematical and Physical Journal for Secondary Schools, New advanced problems: proposed problem A265, April 2001.
L. J. Mordell, On the integer solutions of the equation x^2+y^2+z^2+2xyz=n, J. Lond. Math. Soc. 28 (1953), 500-510.
J. Propp, The combinatorics of frieze patterns and Markoff numbers, arXiv:math/0511633 [math.CO], 2005-2008.
Matthew Christopher Russell, Using experimental mathematics to conjecture and prove theorems in the theory of partitions and commutative and non-commutative recurrences, PhD Dissertation, Mathematics Department, Rutgers University, May 2016.

Cf. A002559, A072878, A072879, A072880.

Markov tree: A327345, A368546.

[n le 3 select 1 else (Self(n-1)^2 + Self(n-2)^2)/Self(n-3): n in [1..16]]; // G. C. Greubel, Nov 07 2024

f:=proc(n) option remember; global K; local i;
if n <= K then 1
else add(f(n-i)^2,i=1..K-1)/f(n-K); fi; end;
K:=3;
[seq(f(n),n=1..10)]; # N. J. A. Sloane, Mar 17 2017

a[n_]:= (a[n-1]^2 +a[n-2]^2)/a[n-3]; a[1]=a[2]=a[3]=1; Array[a, 13] (* Or *)
a[n_]:= 3*a[n-1]*a[n-2] - a[n-3]; a[1]= a[2]= a[3]= 1; Array[a, 13] (* Robert G. Wilson v, Dec 26 2012 *)
nxt[{a_,b_,c_}]:={b,c,(c^2+b^2)/a}; NestList[nxt,{1,1,1},15][[;;,1]] (* Harvey P. Dale, Jul 07 2025 *)

{a(n) = if( n<1, n = 4-n); if( n<4, 1, 3 * a(n-1) * a(n-2) - a(n-3))}; /* Michael Somos, Jan 12 2013 */

{ a=a3=a2=a1=1; for (n = 1, 18, if (n>3, a=(a1^2 + a2^2)/a3; a3=a2; a2=a1; a1=a); write("b064098.txt", n, " ", a) ) } /* Harry J. Smith, Sep 06 2009 */

def A064098(n):
    def a(n): return 1 if n<4 else (a(n-1)^2 + a(n-2)^2)/a(n-3)
    return a(n)
[A064098(n) for n in range(16)] # G. C. Greubel, Nov 07 2024

Conjecture: lim_{n -> infinity} log(log(a(n)))/n exists = 0.48.... - Benoit Cloitre, Aug 07 2002. This is true - see below.
For this subsequence of the Markoff numbers, we have 2^(F(n-1) - 1) < a(n) < 3^(F(n-1) - 1) for n > 4, where F(n) are the Fibonacci numbers with F(0)=0, F(1)=1, F(n+1) = F(n) + F(n-1). Hence log(log(a(n)))/n tends to log((1 + sqrt(5))/2) as previously conjectured. - Andrew Hone, Jan 16 2006
a(n) = 3 * a(n-1) * a(n-2) - a(n-3). a(4-n) = a(n) for all n in Z. - Michael Somos, Jan 12 2013
a(n) ~ 1/3 * c^(((1 + sqrt(5))/2)^n), where c = 1.2807717799265504005186306582930649245... . - Vaclav Kotesovec, May 06 2015

Entry improved by comments from Michael Somos, Sep 25 2001

A072880 A recurrence of order 6: a(1)=a(2)=a(3)=a(4)=a(5)=a(6)=1; a(n) = (a(n-1)^2 + a(n-2)^2 + a(n-3)^2 + a(n-4)^2 + a(n-5)^2)/a(n-6).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 5, 29, 869, 756029, 571580604869, 326704387862983487112029, 21347151409785350408171299054974277225256721769, 15713823217665540462976624783900822313284439536736221766688609460305249837839107387688348185

Offset: 1

Benoit Cloitre, Jul 28 2002

Any six successive terms satisfy the Markoff-Hurwitz equation a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 6*a*b*c*d*e*f. - Bruno Langlois, Aug 09 2016

Seiichi Manyama, Table of n, a(n) for n = 1..17
Andrew N. W. Hone, Diophantine non-integrability of a third order recurrence with the Laurent property, arXiv:math/0601324 [math.NT], 2006.
Andrew N. W. Hone, Diophantine non-integrability of a third order recurrence with the Laurent property, J. Phys. A: Math. Gen. 39 (2006), L171-L177.
Matthew Christopher Russell, Using experimental mathematics to conjecture and prove theorems in the theory of partitions and commutative and non-commutative recurrences, PhD Dissertation, Mathematics Department, Rutgers University, May 2016.

Cf. A006720, A064098, A072878, A072879.

RecurrenceTable[{a[n] == (a[n - 1]^2 + a[n - 2]^2 + a[n - 3]^2 + a[n -  4]^2 + a[n - 5]^2)/a[n - 6], a[1] == a[2] == a[3] == a[4] == a[5] == a[6] == 1}, a, {n, 1, 14}] (* Michael De Vlieger, Aug 11 2016 *)
nxt[{a_, b_, c_, d_, e_, f_}] := {b, c, d, e, f,(b^2+c^2+d^2+e^2+f^2)/a}; NestList[ nxt, Table[1,6],20][[All,1]] (* Harvey P. Dale, Mar 18 2018 *)

a(n) = 6*a(n-1)*a(n-2)*a(n-3)*a(n-4)*a(n-5) - a(n-6). - Bruno Langlois, Aug 09 2016

A022405 a(n) = a(n-1)*a(n-2) - a(n-3), with a(1) = 0, a(2) = 1, a(3) = 2.

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 10, 37, 366, 13532, 4952675, 67019597734, 331926286207224918, 22245566178948766568816183137, 7383888166355511098764350563784314022618210032

Offset: 1

Robert G. Wilson v, Jul 05 2000

Theorem 1.1 of Hare et al. (2010, 2011) involves a shifted version of this sequence and the Fibonacci sequence A000045. (The program by Alonso del Arte below does involve a shifted version of this sequence.) - Petros Hadjicostas, May 11 2019

G. C. Greubel, Table of n, a(n) for n = 1..21
Kevin G. Hare, Ian D. Morris, Nikita Sidorov, and Jacques Theys, An explicit counterexample to the Lagarias-Wang finiteness conjecture, arXiv:1006.2117 [math.OC], 2010-2011.
Kevin G. Hare, Ian D. Morris, Nikita Sidorov, and Jacques Theys, An explicit counterexample to the Lagarias-Wang finiteness conjecture, Advances in Mathematics 226 (2011), 4667-4701.

Cf. A001622, A061021, A061292, A072878, A072879, A072880, A178768.

I:=[0,1,2]; [n le 3 select I[n] else Self(n-1)*Self(n-2) - Self(n-3): n in [1..15]];  // G. C. Greubel, Mar 01 2018

a[1] = 0; a[2] = 1; a[3] = 2; a[n_] := a[n] = a[n - 1] a[n - 2] - a[n - 3]; Table[a[n], {n, 1, 15}] (* Alonso del Arte, Jan 31 2011 *)
nxt[{a_,b_,c_}]:={b,c,c*b-a}; NestList[nxt,{0,1,2},15][[;;,1]] (* Harvey P. Dale, Mar 23 2025 *)

It appears that lim_{n->infinity} log(a(n))/phi^n = 0.07743008049000107520747623421744398272089261907514..., where phi = (1 + sqrt(5))/2 is the golden ratio A001622. - Petros Hadjicostas and Jon E. Schoenfield, May 11 2019

Name clarified by Michel Marcus, May 10 2019

A061292 a(n) = a(n-1)a(n-2)a(n-3) - a(n-4) for n>3 with a(0) = a(1) = a(2) = a(3) = 2.

Original entry on oeis.org

2, 2, 2, 2, 6, 22, 262, 34582, 199330642, 1806032092550706, 12449434806576800059248920402, 4481765860945171681908664776799089162954814190172722

Offset: 0

Stephen G Penrice, Jun 04 2001

Any four consecutive terms are a solution to the Diophantine equation w^2 + x^2 + y^2 + z^2 = wxyz.

a(n) = 2 * A072878(n+1).

Vincenzo Librandi, Table of n, a(n) for n = 0..18
Michael Magee and Brady Haran, A simple equation that behaves weirdly, Numberphile, Youtube video, 2025.

Cf. A022405, A061021, A072878, A072879, A072880.

a061292 n = a061292_list !! n
a061292_list = 2 : 2 : 2 : 2 : zipWith (-)
   (zipWith3 (((*) .) . (*)) (drop 2 xs) (tail xs) xs) a061292_list
   where xs = tail a061292_list
-- Reinhard Zumkeller, Mar 25 2015

I:=[2,2,2,2]; [n le 4 select I[n] else Self(n-1)*Self(n-2)*Self(n-3)-Self(n-4): n in [1..12]]; // Vincenzo Librandi, Sep 17 2011

a[1] := 2; a[2] := 2; a[3] := 2; a[4] := 2; a[n_] := a[n - 1]*a[n - 2]*a[n - 3] - a[n - 4]; Table[a[n], {n, 1, 15}] (* Stefan Steinerberger, Mar 31 2006 *)
RecurrenceTable[{a[0]==a[1]==a[2]==a[3]==2,a[n]==a[n-1]a[n-2]a[n-3]- a[n-4]},a[n],{n,12}] (* Harvey P. Dale, Sep 15 2011 *)

More terms from Larry Reeves (larryr(AT)acm.org) and Jason Earls, Jun 05 2001

A112957 a(1) = a(2) = a(3) = 1; for n > 1, a(n+3) = a(n)^2 + a(n+1)^2 + a(n+2)^2.

Original entry on oeis.org

1, 1, 1, 3, 11, 131, 17291, 298995963, 89398586189293211, 7992107212644486930829797919966571, 63873777698404030240264509605345282496735163325301838600463378485931

Offset: 1

Jonathan Vos Post, Jan 02 2006; definition corrected Jan 02 2006

A quadratic tribonacci sequence.

This is to A000283 as a tribonacci (A000213) is to Fibonacci. Two oddities about this sequence: (a) its first 7 terms are identical to terms numbered 2 through 8 of A072878; (b) only one of the first 9 terms are composite. Primes in the sequence begin 3, 11, 131, 17291 and 89398586189293211. What is the next prime?

Seiichi Manyama, Table of n, a(n) for n = 1..14

Cf. A000213, A000283, A072878, A112958, A112959, A112960.

Join[{a=1,b=1,c=1},Table[d=a*a+b*b+c*c;a=b;b=c;c=d,{n,10}]] (* Vladimir Joseph Stephan Orlovsky, Apr 19 2011 *)

A072877 a(1) = a(2) = a(3) = a(4) = 1; a(n) = (a(n-1)*a(n-3) + a(n-2)^4)/a(n-4).

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 19, 119, 65339, 67258454, 959259994615659593, 171965197021698738644442682357, 12959040525296547835480490169418622922155526267774117749963303914461

Offset: 1

Benoit Cloitre, Jul 28 2002

A variation of a Somos-4 sequence with a(n-2)^4 in place of a(n-2)^2.

Seiichi Manyama, Table of n, a(n) for n = 1..17
Joshua Alman, Cesar Cuenca, and Jiaoyang Huang, Laurent phenomenon sequences, Journal of Algebraic Combinatorics 43(3) (2015), pp. 589-633.
S. Fomin and A. Zelevinsky, The Laurent Phenomenon, Advances in Applied Mathematics 28 (2002), pp. 119-144.
David Gale, The strange and surprising saga of the Somos sequences, Math. Intelligencer 13(1) (1991), pp. 40-42.
Andrew N. W. Hone, Diophantine non-integrability of a third order recurrence with the Laurent property, arXiv:math/0601324 [math.NT], 2006.
Andrew N. W. Hone, Diophantine non-integrability of a third order recurrence with the Laurent property, J. Phys. A: Math. Gen. 39 (2006), pp. L171-L177.

Cf. A006720, A022405, A061292, A065918, A072878, A072879, A072880, A074394, A178768.

L[0]:=0; L[1]:=0; L[2]:=0; L[3]:=0; for n from 0 to 4000 do L[n+4]:=evalf(ln(1+exp(L[n+3]+L[n+1]-4*L[n+2]))+4*L[n+2]-L[n]): od: for n from 3990 to 4000 do print(evalf(ln(L[n+4])/(n+4))): od: #Note: L[n] is log(a[n]) # Andrew Hone, Nov 15 2005

nxt[{a_,b_,c_,d_}]:={b,c,d,(d*b+c^4)/a}; NestList[nxt,{1,1,1,1},15][[All,1]] (* Harvey P. Dale, Jun 01 2022 *)

Lim_{n->infinity} (log(log(a(n))))/n = log(2+sqrt(3))/2 = A065918/2 or about 0.658. - Andrew Hone, Nov 15 2005; corrected by Michel Marcus, May 12 2019
From Jon E. Schoenfield, May 12 2019: (Start)
It appears that, for n >= 1,
  a(n) = ceiling(e^(c0*x^n + d0/x^n)) if n is even,
         ceiling(e^(c1*x^n + d1/x^n)) if n is odd,
where
  x  = sqrt(2 + sqrt(3)) = (sqrt(2) + sqrt(6))/2
  c0 =   0.024915247166055931001426396817534982995670642690...
  c1 =   0.029604794868229453467890216788323427656809346011...
  d0 = -10.535089427608481105514469573411011428431309483956...
  d1 =  -2.856773870202800001336732759121362374871088274450...
(End)

Definition corrected by Matthew C. Russell, Apr 24 2012

A276095 A nonlinear recurrence of order 4: a(1)=a(2)=a(3)=a(4)=1; a(n)=(a(n-1)+a(n-2)+a(n-3))^2/a(n-4).

Original entry on oeis.org

1, 1, 1, 1, 9, 121, 17161, 298978681, 9933176210033041, 815437979830770470704295274609, 38747106750801481775941360512378545527545442200632960401

Offset: 1

Seiichi Manyama, Aug 18 2016

nonn

All terms are perfect squares.

Seiichi Manyama, Table of n, a(n) for n = 1..15

Cf. A072878, A072882.

RecurrenceTable[{a[n] == (a[n - 1] + a[n - 2] + a[n - 3])^2/a[n - 4], a[1] == a[2] == a[3] == a[4] == 1}, a, {n, 1, 12}] (* Michael De Vlieger, Aug 18 2016 *)
nxt[{a_,b_,c_,d_}]:={b,c,d,(b+c+d)^2/a}; NestList[nxt,{1,1,1,1},10][[;;,1]] (* Harvey P. Dale, Aug 20 2024 *)

def A(m, n)
  a = Array.new(m, 1)
  ary = [1]
  while ary.size < n
    i = a[1..-1].inject(:+)
    j = i * i
    break if j % a[0] > 0
    a = *a[1..-1], j / a[0]
    ary << a[0]
  end
  ary
end
def A276095(n)
  A(4, n)
end

a(n) = A072878(n)^2.
a(n) = 16*a(n-1)*a(n-2)*a(n-3) - 2a(n-1) - 2a(n-2) - 2a(n-3) - a(n-4).
a(n)*a(n-1)*a(n-2)*a(n-3) = ((a(n) + a(n-1) + a(n-2) + a(n-3))/4)^2.

A061021 a(n) = a(n-1)*a(n-2) - a(n-3) with a(0) = a(1) = a(2) = 3.

Original entry on oeis.org

3, 3, 3, 6, 15, 87, 1299, 112998, 146784315, 16586334025071, 2434613678231239448367, 40381315689150066251526220641224742, 98312903521778500654864668915856114278134197773017871243

Offset: 0

Stephen G Penrice, May 23 2001

Any three consecutive terms are a solution to the Diophantine equation x^2 + y^2 + z^2 = xyz.

Harry J. Smith, Table of n, a(n) for n = 0..17
Loren C. Larson, Solution to Problem Proposal 701, College Mathematics Journal 33 (2002), pp. 241-242.
Edward T. H. Wang, Problem Proposal 701, College Mathematics Journal 32 (2001), p. 211.

Cf. A022405, A061292, A072878, A072879, A072880, A074394, A178768.

a061021 n = a061021_list !! n
a061021_list = 3 : 3 : 3 : zipWith (-)
(tail $ zipWith (*) (tail a061021_list) a061021_list) a061021_list
-- Reinhard Zumkeller, Mar 25 2015

RecurrenceTable[{a[n] == a[n - 1] a[n - 2] - a[n - 3], a[0] == a[1] == a[2] == 3}, a, {n, 0, 12}] (* Michael De Vlieger, Aug 21 2016 *)

for (n=0, 17, if (n>2, a=a1*a2 - a3; a3=a2; a2=a1; a1=a, if (n==0, a=a3=3, if (n==1, a=a2=3, a=a1=3))); write("b061021.txt", n, " ", a)) \\ Harry J. Smith, Jul 16 2009

From Jon E. Schoenfield, May 12 2019: (Start)
It appears that, for n >= 1,
  a(n) = ceiling(e^(c0*phi^n - c1/(-phi)^n))
where
  phi = (1 + sqrt(5))/2,
   c0 = 0.4004033011137849744572073756789830081726425559860...
   c1 = 0.2798639753144007577581523025628820390768226527315...
(End)

More terms from Erich Friedman, Jun 03 2001

Name clarified by Petros Hadjicostas, May 11 2019

A074394 a(n) = a(n-1)*a(n-2) - a(n-3) with a(1) = 1, a(2) = 2, and a(3) = 3.

Original entry on oeis.org

1, 2, 3, 5, 13, 62, 801, 49649, 39768787, 1974480504962, 78522694637486171445, 155041529758800625329015665441303, 12174278697379026530632791354719900462885271361687873

Offset: 1

Henry Bottomley, Sep 24 2002

nonn

All consecutive quadruples are pairwise coprime. Multiples of 2 occur when n=2 mod 4, multiples of 3 when n=3 mod 4, multiples of 5 when n=4 mod 7, multiples of 7 when n=10 mod 14, multiples of 9 when n=7 or 11 mod 24, multiples of 10 when n=18 mod 28. Multiples of 4, 6 and 8 never occur.

			a(6) = a(5)*a(4) - a(3) = 13*5 - 3 = 62.

Reinhard Zumkeller, Table of n, a(n) for n = 1..19

Cf. A001622, A022405, A061021, A072878, A072879, A072880.

a074394 n = a074394_list !! (n-1)
a074394_list = 1 : 2 : 3 : zipWith (-)
   (tail $ zipWith (*) (tail a074394_list) a074394_list) a074394_list
-- Reinhard Zumkeller, Mar 25 2015

nxt[{a_,b_,c_}]:={b,c,b*c-a}; NestList[nxt,{1,2,3},15][[All,1]] (* Harvey P. Dale, Jan 21 2023 *)

Lim_{n->infinity} a(n+1)/a(n)^phi = 1, where phi is the golden ratio (1+sqrt(5))/2 = A001622. - Benoit Cloitre, Sep 26 2002
From Jon E. Schoenfield, May 13 2019: (Start)
It appears that, for n >= 2,
  a(n) = ceiling(e^(c*phi^n - d/(-phi)^n))
where
  phi = (1 + sqrt(5))/2
    c = 0.230193077518834725477008740044380256486365499661...
    d = 0.067704372842879037264190305626317036100889750046...
(End)

cp's OEIS Frontend

A072879 a(n) = 5*a(n-1)*a(n-2)*a(n-3)*a(n-4) - a(n-5) with a(1) = a(2) = a(3) = a(4) = a(5) = 1.

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A064098 a(n+1) = (a(n)^2 + a(n-1)^2)/a(n-2), with a(1) = a(2) = a(3) = 1.

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A072880 A recurrence of order 6: a(1)=a(2)=a(3)=a(4)=a(5)=a(6)=1; a(n) = (a(n-1)^2 + a(n-2)^2 + a(n-3)^2 + a(n-4)^2 + a(n-5)^2)/a(n-6).

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A022405 a(n) = a(n-1)*a(n-2) - a(n-3), with a(1) = 0, a(2) = 1, a(3) = 2.

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A061292 a(n) = a(n-1)*a(n-2)*a(n-3) - a(n-4) for n>3 with a(0) = a(1) = a(2) = a(3) = 2.

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A112957 a(1) = a(2) = a(3) = 1; for n > 1, a(n+3) = a(n)^2 + a(n+1)^2 + a(n+2)^2.

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A072877 a(1) = a(2) = a(3) = a(4) = 1; a(n) = (a(n-1)*a(n-3) + a(n-2)^4)/a(n-4).

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A072879 a(n) = 5a(n-1)a(n-2)a(n-3)a(n-4) - a(n-5) with a(1) = a(2) = a(3) = a(4) = a(5) = 1.

A061292 a(n) = a(n-1)a(n-2)a(n-3) - a(n-4) for n>3 with a(0) = a(1) = a(2) = a(3) = 2.